Question

✌ don't spam. ✌. answer with reason​

Answer 1

Given,

$\displaystyle\to f(x)=\lim_{n\to\infty}\left(\dfrac{n^n(x+n)\left(x+\dfrac{n}{2}\right)\,\dots\,\left(x+\dfrac{n}{n}\right)}{n!\left(x^2+n^2\right)\left(x^2+\dfrac{n^2}{4}\right)\,\dots\,\left(x^2+\dfrac{n^2}{n^2}\right)}\right)^{\dfrac{x}{n}}$

$\displaystyle\to f(x)=\lim_{n\to\infty}\left(\dfrac{n^n\displaystyle\prod_{r=1}^n\left(x+\dfrac{n}{r}\right)}{\displaystyle n!\prod_{r=1}^n\left(x^2+\dfrac{n^2}{r^2}\right)}\right)^{\dfrac{x}{n}}$

$\displaystyle\to f(x)=\lim_{n\to\infty}\left(\dfrac{n^n\displaystyle\prod_{r=1}^nn\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle n!\prod_{r=1}^nn^2\left(\dfrac{x^2}{n^2}+\dfrac{1}{r^2}\right)}\right)^{\dfrac{x}{n}}$

$\displaystyle\to f(x)=\lim_{n\to\infty}\left(\dfrac{n^n\displaystyle\prod_{r=1}^nn\cdot\prod_{r=1}^n\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle n!\prod_{r=1}^nn^2\cdot\prod_{r=1}^n\left(\dfrac{x^2}{n^2}+\dfrac{1}{r^2}\right)}\right)^{\dfrac{x}{n}}$

$\displaystyle\to f(x)=\lim_{n\to\infty}\left(\dfrac{n^n\cdot n^n\cdot\displaystyle\prod_{r=1}^n\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle n!\cdot(n^2)^n\cdot\prod_{r=1}^n\left(\dfrac{x^2}{n^2}+\dfrac{1}{r^2}\right)}\right)^{\dfrac{x}{n}}$

$\displaystyle\to f(x)=\lim_{n\to\infty}\left(\dfrac{n^{2n}\cdot\displaystyle\prod_{r=1}^n\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle n!\cdot n^{2n}\cdot\prod_{r=1}^n\left(\dfrac{x^2}{n^2}+\dfrac{1}{r^2}\right)}\right)^{\dfrac{x}{n}}$

$\displaystyle\to f(x)=\lim_{n\to\infty}\left(\dfrac{\displaystyle\prod_{r=1}^n\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle n!\cdot\prod_{r=1}^n\left(\dfrac{x^2}{n^2}+\dfrac{1}{r^2}\right)}\right)^{\dfrac{x}{n}}$

Note that $f(x)>0\quad\!\forall\,x>0.$

Taking logarithm,

$\displaystyle\to\log f(x)=\lim_{n\to\infty}\left(\dfrac{x}{n}\right)\log\left(\dfrac{\displaystyle\prod_{r=1}^n\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle n!\cdot\prod_{r=1}^n\left(\dfrac{x^2}{n^2}+\dfrac{1}{r^2}\right)}\right)$

$\to\log f(x)=\displaystyle\lim_{n\to\infty}\left(\dfrac{x}{n}\right)\log\left(\dfrac{\displaystyle\prod_{r=1}^n\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle \prod_{r=1}^nr\cdot\prod_{r=1}^n\left(\dfrac{x^2}{n^2}+\dfrac{1}{r^2}\right)}\right)$

$\to\log f(x)=\displaystyle\lim_{n\to\infty}\left(\dfrac{x}{n}\right)\log\left(\dfrac{\displaystyle\prod_{r=1}^n\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle\prod_{r=1}^nr\left(\dfrac{x^2}{n^2}+\dfrac{1}{r^2}\right)}\right)$

$\to\log f(x)=\displaystyle\lim_{n\to\infty}\left(\dfrac{x}{n}\right)\log\left(\dfrac{\displaystyle\prod_{r=1}^n\left(\dfrac{x}{n}+\dfrac{1}{r}\right)}{\displaystyle\prod_{r=1}^n\left(\dfrac{x^2r}{n^2}+\dfrac{1}{r}\right)}\right)$

$\to\log f(x)=\displaystyle\lim_{n\to\infty}\left(\dfrac{x}{n}\right)\log\left(\dfrac{\displaystyle\prod_{r=1}^n\dfrac{1}{r}\left(\dfrac{xr}{n}+1\right)}{\displaystyle\prod_{r=1}^n\dfrac{1}{r}\left(\dfrac{x^2r^2}{n^2}+1\right)}\right)$

$\to\log f(x)=\displaystyle\lim_{n\to\infty}\left(\dfrac{x}{n}\right)\log\left(\dfrac{\displaystyle\prod_{r=1}^n\left(\dfrac{xr}{n}+1\right)}{\displaystyle\prod_{r=1}^n\left(\dfrac{x^2r^2}{n^2}+1\right)}\right)$

Since $\displaystyle\log\left(\prod_{r=a}^bx_r\right)=\sum_{r=a}^b\log(x_r),$

$\to\log f(x)=\displaystyle\lim_{n\to\infty}\left(\dfrac{x}{n}\right)\sum_{r=1}^n\left[\log\left(\dfrac{xr}{n}+1\right)-\log\left(\dfrac{x^2r^2}{n^2}+1\right)\right]$

$\to\log f(x)=\displaystyle\lim_{n\to\infty}\sum_{r=1}^n\log\left(\dfrac{xr}{n}+1\right)\cdot\left(\dfrac{x}{n}\right)-\lim_{n\to\infty}\sum_{r=1}^n\log\left(\dfrac{x^2r^2}{n^2}+1\right)\cdot\left(\dfrac{x}{n}\right)$

We can change the sums to integrals as:

• $\displaystyle\lim_{n\to\infty}\sum_{r=0}^nf\left(\dfrac{br}{n}\right)\cdot\dfrac{b}{n}=\int\limits_0^bf(x)\ dx$

Thus we get,

$\to\log f(x)=\displaystyle\int\limits_0^x\log(x+1)\ dx-\int\limits_0^x\log(x^2+1)\ dx$

$\to\log f(x)=\displaystyle\int\limits_0^x\log\left(\dfrac{x+1}{x^2+1}\right)\ dx$

Differentiating,

$\to\dfrac{f'(x)}{f(x)}=\log\left(\dfrac{x+1}{x^2+1}\right)$

Let $x=2.$

$\to\dfrac{f'(2)}{f(2)}=\log\left(\dfrac{3}{5}\right)<0$

$\implies\underline{\underline{f'(2)\leq0}}$

Let $x=3.$

$\to\dfrac{f'(3)}{f(3)}=\log\left(\dfrac{2}{5}\right)<\log\left(\dfrac{3}{5}\right)=\dfrac{f'(2)}{f(2)}$

Here $f'(x)>0$ for $x\in[0,\ 1].$ It means $f(x)$ is increasing in this interval.

Therefore,

$\to f\left(\dfrac{1}{2}\right)\leq f(1)$

$\to\underline{\underline{f\left(\dfrac{1}{3}\right)\leq f\left(\dfrac{2}{3}\right)}}$

Hence answers are (B) and (C).